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Problem: A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the x-y plane with centre at O and constant angular speed $\omega$. If the angular momentum of the system, calculated about O and P are denoted by $\vec{L}_{O}$ and $\vec{L}_{P}$, respectively, then (IIT JEE 2012)
Solution: For a particle of mass $m$ having position $\vec{r}$ and velocity $\vec{v}$, the angular momentum $\vec{L}$ about a point Q is given by \begin{align} \vec{L}=m(\vec{r}-\vec{r}_Q)\times \vec{v},\nonumber \end{align} where $\vec{r}_Q$ is the position vector of the point Q.
Consider the Cartesian and polar coordinate systems as shown in the figure. Let $h$ be the height of the point P above the point O. The angular momenta of the system about the point O and about the point P are \begin{align} \vec{L}_O&=m(\vec{r}-\vec{r}_{O})\times\vec{v} \\ &=m\,r\vec{e}_r\times \omega r\vec{e}_t \\ &=m\omega r^2\,\hat{z},\nonumber\\ \vec{L}_P&=m(\vec{r}-\vec{r}_P)\times\vec{v} \\ & =m\,(r\vec{e}_r -h\hat{z})\times \omega r\vec{e}_t\nonumber\\ &=mh\omega r\vec{e}_r+m\omega r^2\,\hat{z}.\nonumber \end{align} Since direction $\vec{e}_r$ rotates with time, $\vec{L}_{P}$ varies with time. However, $\vec{L}_O$ and $|\vec{L}_P|$ do not vary with time. I encourage you to deduce the results by finding the torque about O and P. \emph{Hint:} $\vec{\tau}_O=\vec{0}$ and $\vec{\tau}_P\neq \vec{0}$.